Emirp numbers | C # program to print Emirp numbers less than n ; Function to find reverse of any number ; Sieve method used for generating emirp number ( use of sieve of Eratosthenes ) ; Create a boolean array " prime [ 0 . . n ] " and initialize all entries it as true . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Traverse all prime numbers ; Find reverse a number ; A number is emrip if it is not a palindrome number and its reverse is also prime . ; Mark reverse prime as false so that it 's not printed again ; Driver code
using System ; class GFG { static int reverse ( int x ) { int rev = 0 ; while ( x > 0 ) { rev = ( rev * 10 ) + x % 10 ; x = x / 10 ; } return rev ; } static void printEmirp ( int n ) { bool [ ] prime = new bool [ n + 1 ] ; for ( int i = 0 ; i < n + 1 ; i ++ ) prime [ i ] = true ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = false ; } } for ( int p = 2 ; p <= n ; p ++ ) { if ( prime [ p ] ) { int rev = reverse ( p ) ; if ( p != rev && rev <= n && prime [ rev ] ) { Console . Write ( p + " ▁ " + rev + " ▁ " ) ; prime [ rev ] = false ; } } } } public static void Main ( ) { int n = 100 ; printEmirp ( n ) ; } }